I think I solved it. I want someone to check it.
$f(x)=\frac{1}{(1+x)^2}+e^x$ around $c=2$
$t=x-2$
$f(t)=\frac{1}{(t+3)^2}+e^{t+2}$
$f(t)=\frac{1}{9(1-(-\frac{t}{3}))^2}+e^{t+2}$
For $|\frac{-t}{3}|<1$
$f(t)=\frac{1}{9}\sum_{n=1}^\infty n(\frac{-t}{3})^{n-1}+\sum_{n=0}^\infty\frac{(t+2)^n}{n!}$
$f(t)=\frac{1}{9}\sum_{n=1}^\infty n(\frac{-t}{3})^{n-1}+\sum_{n=1}^\infty\frac{(t+2)^n}{n!}+1$
$f(t)=\sum_{n=1}^\infty (n)((-1)^{n-1})((\frac{1}{3})^{n+1})(t^{n-1})+\sum_{n=1}^\infty\frac{(t+2)^n}{n!}+1$
$f(t)=\sum_{n=1}^{\infty}[(n)((-1)^{n-1})((\frac{1}{3})^{n+1})(t^{n-1})+\frac{(t+2)^n}{n!}]+1$
$f(x)=\sum_{n=1}^{\infty}[(n)((-1)^{n-1})((\frac{1}{3})^{n+1})((x-2)^{n-1})+\frac{(x)^n}{n!}]+1$
For $|t|<3$, so $x\in\langle-1,5\rangle$. Radius of convergence is $3$.